Question

我尝试进行查询，以显示所有工作时间，日期和人员。我有那种奔跑。

我有一张名为uren的表：

rec_id | user_id | start (datetime) | eind (datetime)

我有一个名为users的表

user_id | name        |

通过下面的查询，我几乎拥有了我想要的所有信息。

select users.name, sec_to_time(SUM(TIME_TO_SEC(TIMEDIFF(uren.eind, uren.start)))),count(distinct(date(start))) as dagen
from uren, users
where date(uren.start) between CAST('2017-10-04 00:00:00' as Date) and CAST('2017-11-04 00:00:00' as DATE) and
      uren.user_id = users.user_id
group by uren.user_id
ORDER BY name

向我展示了这个

Piet (name) 230 (hours total) 24(days worked)

现在出现了真正的问题：

我希望每工作一天少于5小时减去30分钟。

我无能为力。请帮助

Answer 1

假设每个用户每天一行：

select users.name,
       (sec_to_time(sum(time_to_sec(timediff(uren.eind, uren.start))) -
                    30 * 60 * sum(time_to_sec(timediff(uren.eind, uren.start)) < 5*60*60)
                   )
       ),
       count(distinct(date(start))) as dagen
from uren join users
     on uren.user_id = users.user_id
where date(uren.start) between '2017-10-04' and '2017-11-04'   
group by uren.user_id
order by name;

如果有一天有多班，您需要先按天聚合：

select u.name,
       (sec_to_time(sum(day_secs) -
                    30 * 60 * sum(day_secs < 5*60*60)
                   )
       ),
       count(*) as dagen
from (select uren.user_id, uren.name, date(uren.start),
             sum(time_to_sec(timediff(uren.eind, uren.start))) as day_secs
      from uren join
           users
           on uren.user_id = users.user_id
      where uren.start >= '2017-10-04' and uren.start < '2017-11-05'   
      group by uren.user_id, date(uren.start)
     ) u
group by name
order by name

Answer 2

有一个很好的理由，你没有线索。这是因为你所要求的是相当复杂的。

你需要接受的第一件事（我的意思是绝望地接受）是你的功能取决于天和用户，因此，你需要两者一组，一开始，然后只有最终结果的用户。为此，您需要一个按天和用户分组的子查询，然后父母可以按用户进行分组。

这是我想出的......

SELECT 
users.name,
sec_to_time(SUM(ur.timeWorked)) as tWorked,
SUM(ur.dagen) as Dagen

 FROM users
 INNER JOIN ( 

   SELECT
   user_id,

   SUM(TIME_TO_SEC(TIMEDIFF(`eind`, `start`))) - (1800 * 
     IF(SUM(TIME_TO_SEC(TIMEDIFF(`eind`, `start`))) < 18000,1,0))
   as `timeWorked`,

  count(distinct(date(`start`))) as `dagen`

  FROM uren

  WHERE date(`start`) 
    BETWEEN CAST('2017-10-04 00:00:00' as Date) 
    AND CAST('2017-11-04 00:00:00' as DATE)
      GROUP BY user_id, DAYOFYEAR(`start`)

    ) as ur ON ur.user_id = users.user_id

GROUP BY ur.user_id
ORDER BY name

如何在mysql

2 个答案: